What is the sum from n=1 to infinity of 5/(n^2+16) ?

The sum from n=1 to infinity of 5/(n^2+16) is converges

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sum from n=1 to infinity of 5/(n^2+16)

n = 1 \sum \infty \frac{5}{n ^{2} + 16}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 16}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 16}

Apply Series Comparison Test:

converges

= 5 converges

= converges

s^{2} + 6 s + 7

x \to \frac{π}{2} lim (x)

y = \frac{7 ln ( x )}{3 x + 4}

x \to \frac{1}{3} lim (3 x^{3} + 2 x^{2})

y = \frac{x}{x ^{2} + 1}, at x = - 1

What is the sum from n=1 to infinity of 5/(n^2+16) ?
The sum from n=1 to infinity of 5/(n^2+16) is converges